Revisiting Fracture Mechanic Handbooks
with (Interpretable) Sparse Regression
MePhy & I-Gaia Day: Machine Learning in mechanics and physics
December 9, 2025.
Linear Elastic Fracture Mechanics
Source : https://en.wikipedia.org/wiki/Fracture_mechanics.
Typical problem
An elastic body \(\Omega\) with a crack \(\Gamma\) under a given load.
Assumption : Non-linearities are confined close to the crack tip.
Question : Does the crack propagates ?
Crack tip singularity
Stress field (Williams, 1957)
\[ \sigma_{ij}(r, \theta) = \frac{\color{red}{K_I}}{\sqrt{2\pi r}} f_{ij}^{I}(\theta) + \frac{\color{red}{K_{II}}}{\sqrt{2\pi r}} f_{ij}^{II}(\theta) + \frac{\color{red}{K_{III}}}{\sqrt{2\pi r}} f_{ij}^{III}(\theta) + \mathcal{O}(1), \quad \color{gray}{\sigma_{ij}(r, \theta) \underset{r \to 0}{\to} \infty.} \]
There is a stress singularity at crack tip.
Stress Intensity Factors (SIFs)
Denoted \(K_p\), they :
- represent the crack tip sollicitation,
- depend on geometry and load.
Designing structures with cracks
Crack propagation criteria
Propagation criteria rely on the SIFs \(K_p\), e.g.,
\(\quad \displaystyle \color{red}{K_I} \leq K_{Ic}\), or
\(\quad \displaystyle G = \frac{\color{red}{K_{I}}^2 + \color{red}{K_{II}}^2}{E'} + \frac{\color{red}{K_{III}}^2}{2 \mu} \leq G_c\).
How to determine the SIFs ?
Different methods are employed :
- analytical (e.g., crack in (semi-)infinite domain TODO: Change to a specific method),
- numerical (e.g., FEM simulations),
- experimental (e.g., photo-elasticity, DIC.).
Known solutions have been compiled in handbooks!
Fracture mechanic design handbooks
(Tada et al., 2000)
Example (Tada et al., 2000)
Methodology
- Dimensional analysis to find relevant parameters
- dimensionless function \(F(\color{green}{\mathbf{p}})\),
depending on parameters \(\color{green}{\mathbf{p}}\) (geometry & load).
- dimensionless function \(F(\color{green}{\mathbf{p}})\),
- Find \(F(\color{green}{\mathbf{p}})\) as truncated series expansion, or
fit of numerical/experimental data, or …
Limitations
Dense polynomials, limited to few parameters, etc.
Ex. 1 : CT specimen - Setup & Data
Compact Tension (CT) specimen
Sparse regression setup
| Observations | \(a\), \(K_I\) (from numerical simulation) |
| Parameters | \(a\), \(\exp(a)\), \(\frac{1}{\sqrt{b-a}}\) |
| Output | \(\hat{K}_I = \frac{K_I}{\sigma \sqrt{a}}\) with \(\sigma=P/b\) |
Observations
Ex. 1 : CT specimen - Tradeoff - Deg=2
Code output
=== Model 1
Assessment :
- Hyperparameter beta : 3.83e+02
- Number of coefficients : 1
- R2 score : 0.000
LaTeX expression:
\hat{K}_{I}(a) =
+45.6
=== Model 2
Assessment :
- Hyperparameter beta : 1.02e+01
- Number of coefficients : 2
- R2 score : 0.967
LaTeX expression:
\hat{K}_{I}(a) =
-21.1
+17.9 \frac{1}{\sqrt{b-a}}^2
=== Model 3
Assessment :
- Hyperparameter beta : 5.11e-01
- Number of coefficients : 3
- R2 score : 0.997
LaTeX expression:
\hat{K}_{I}(a) =
+5.43
-14.4 \exp(a/b)^2
+22.9 \frac{1}{\sqrt{b-a}}^2Ex. 1 : CT specimen - Tradeoff - Deg=3
Code output
=== Model 1
Assessment :
- Hyperparameter beta : 1.79e+03
- Number of coefficients : 1
- R2 score : 0.000
LaTeX expression:
\hat{K}_{I}(a) =
+45.6
=== Model 2
Assessment :
- Hyperparameter beta : 3.43e+00
- Number of coefficients : 2
- R2 score : 0.999
LaTeX expression:
\hat{K}_{I}(a) =
+3.75
+4.11 \frac{1}{\sqrt{b-a}}^3
=== Model 3
Assessment :
- Hyperparameter beta : 6.13e-01
- Number of coefficients : 3
- R2 score : 1.000
LaTeX expression:
\hat{K}_{I}(a) =
+5.32
-0.383 \exp(a/b)^3
+4.19 \frac{1}{\sqrt{b-a}}^3Ex. 2 : Rotating structure - Setup & Data
Experimental setup
By G. Eliyahu-Yakir and P. Reis (EPFL)
Plate geometry
Illustration
Ex. 2 : Rotating structure - Physics
Illustration
7-parameter problem
- Geometry: \(\color{green}{W, L, a, L_c}\).
- 3D volumic loads:
- centrifugal : \(\boldsymbol{f}_\Omega(r, \theta) \propto \color{green}{\Omega}\)
- along \(\boldsymbol{e}_r\) and \(\boldsymbol{e}_\theta\),
- Euler : \(\boldsymbol{f}_\dot{\Omega}(r, \theta) \propto \color{green}{\dot{\Omega}}\)
- along \(\boldsymbol{e}_2 = \cos(\color{green}{\alpha}) \boldsymbol{e}_\theta - \sin(\color{green}{\alpha}) \boldsymbol{e}_z\).
- centrifugal : \(\boldsymbol{f}_\Omega(r, \theta) \propto \color{green}{\Omega}\)
Dimensional analysis (towards a 3-parameter problem)
The superposition principle (\(K_p = K_{p}^{\Omega} + K_{p}^{\dot{\Omega}}\)) and the Buckingham \(\Pi\) theorem give (\(\frac{L}{W}\) is fixed)
- centrifugal load
- \(K_{I}^{\Omega} = \rho \color{green}{\Omega}^2 \color{green}{W}^{5/2} \color{red}{\hat{K}_{I}^{\Omega}}\left(\color{green}{\frac{a}{W}}, \color{green}{\frac{L_c}{L}}, \color{gray}{\frac{L}{W}}\right)\)
- \(K_{II}^{\Omega} = \rho \color{green}{\Omega}^2 \color{green}{W}^{5/2} \color{red}{\hat{K}_{II}^{\Omega}}\left(\color{green}{\frac{a}{W}}, \color{green}{\frac{L_c}{L}}, \color{gray}{\frac{L}{W}}\right)\)
- \(K_{III}^{\Omega} = 0\),
- Euler load
- \(K_{I}^{\dot{\Omega}} = \rho \color{green}{\dot{\Omega}} \color{green}{W}^{5/2} \cos(\color{green}{\alpha}) \color{red}{\hat{K}_{I}^{\dot{\Omega}}}\left(\color{green}{\frac{a}{W}}, \color{green}{\frac{L_c}{L}}, \color{gray}{\frac{L}{W}}\right)\)
- \(K_{II}^{\dot{\Omega}} = \rho \color{green}{\dot{\Omega}} \color{green}{W}^{5/2} \cos(\color{green}{\alpha}) \color{red}{\hat{K}_{II}^{\dot{\Omega}}}\left(\color{green}{\frac{a}{W}}, \color{green}{\frac{L_c}{L}}, \color{gray}{\frac{L}{W}}\right)\)
- \(K_{III}^{\dot{\Omega}} = \rho \color{green}{\dot{\Omega}} \color{green}{W}^{5/2} \sin(\color{green}{\alpha}) \color{red}{\hat{K}_{III}^{\dot{\Omega}}}\left(\color{green}{\frac{a}{W}}, \color{green}{\frac{L_c}{L}}, \color{gray}{\frac{L}{W}}\right)\).
Ex. 2 : Rotating structure - Data
Data generation
Elastic FEM simulations
Ex. 2 : Rotating structure - Model \(\hat{K}_{I}^{\dot{\Omega}}\)
Thanks for your attention !
F. Loiseau, Y. Epongue Djeugoue, G. Eliyahu-Yakir, P. Reis, V. Lazarus.
Presentation
Ce travail est financé par l’Agence de l’Innovation de Défense – AID – via le Centre Interdisciplinaire d’Etudes pour la Défense et la Sécurité – CIEDS – (projects 2022 - FracAddi).